How Does a Red Giant Star Produce Energy Through Fusion Processes?

• colloio
In summary, the conversation discusses the energy production and composition of a red giant star. The mean density and chemical composition are assumed for the hydrogen shell source and the core. The energy production rates for relevant processes are given. It is assumed that the energy production per gram is identical for central triple-alpha fusion and shell source CNO fusion. The relative mass loss for hydrogen and helium-to-carbon fusion is also mentioned. The questions asked pertain to calculating the energy production rate and temperature in the hydrogen shell source and helium burning core.
colloio
Exercise_C_01
We consider a red giant star. The energy is produced by hydrogen fusion in a shell
and by helium fusion in the core.
We assume that the mean density in the hydrogen shell source is 30 g/cm3 and that
the mean chemical composition is X=0.35, Y=0.63 and Z=0.02.
The mean density in the core is assumed to be 6000 g/cm3 and the mean
composition is Y=0.49 and Z=0.51.
The energy production rates for the relevant processes are:

εPP_I=9*10^-6*X^2*(ρ/(g*cm^3)*(T/(10^6*T)^4 erg/s/g

εCNO=1.8*10^-21*X*Z*(ρ/(g*cm^3)*(T/(10^6*T)^18 erg/s/g

ε=1.7*10^-67*Y^3*(ρ/(g*cm^3)^2*(T/(10^6*T)^30 erg/s/g

We assume ε3α(centre)= εCNO(shell source) i.e. the energy production per gram
material is identical for the central triple-alpha fusion and the shell source CNOfusion.
The relative mass loss for hydrogen fusion is 0.7% and for helium-to-carbon fusion is
0.07%. We now assume that the star will use 2 million years to transform all the
helium to carbon in the core. We also assume that the star will have Y=0.98 at the
beginning of Helium-to-carbon fusion and that the energy production rate is constant
throughout the helium burning.C_01_1: Show that ε3α ≈ 10000 erg/g/s (using the above assumptions).
C_01_2: Calculate the temperature in the hydrogen shell source and in the
helium burning core. Show that CNO is dominating the hydrogen fusion and
that the PP-fusion rate is small.

I have tryed C_01_1 with E=m*c^2 without luck i get 97 erg/s/g and i have tried every kinda way to get 9700 instead, but no, i can't see it.

C_01_2 I am totally lost here don't know how to find the tempratur without ε for hydrogen.

colloio said:
I have tryed C_01_1 with E=m*c^2 without luck i get 97 erg/s/g and i have tried every kinda way to get 9700 instead, but no, i can't see it.
How can we know what went wrong if you don't show your work?

You should define X, Y and Z, but I can guess that they are the fraction of hydrogen, helium and metals (by number of atoms or by weight?).

T/(10^6*T) is T/(10^6*K)?

What is energy production per gram?

Energy production per gram is the amount of energy produced per unit of mass. It is a measure of how efficient a substance or system is at converting its mass into usable energy.

Why is energy production per gram important?

Energy production per gram is important because it helps us understand the efficiency and potential of different energy sources. It also allows us to compare and evaluate different energy production methods.

How is energy production per gram calculated?

Energy production per gram is calculated by dividing the total amount of energy produced by the mass of the substance or system. This can be expressed in various units such as joules per gram or kilowatt-hours per gram.

What factors affect energy production per gram?

The main factors that affect energy production per gram include the type of energy source, the conversion process, and the efficiency of the system. Other factors may also include the purity and density of the substance, as well as external factors such as temperature and pressure.

How can we improve energy production per gram?

There are several ways to improve energy production per gram, such as using more efficient conversion processes, increasing the purity and density of the substance, and implementing technology advancements. Additionally, investing in research and development can help us discover new and more efficient energy sources.

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